B. I think is the correct answer
fraction equation is<span>
F =µR
F=friction,µ=coefficient , R=reaction = mg
use same equation for b part, but the reaction is no longer mg because the plain is now inclined. Draw a forces diagram and you will see that the reaction force can be calculated from the weight of the object and inclination of the plain using trigonometry.</span>
Given : Time taken to reach the maximum height t=3 s a=−g=−10m/s
2
The initial velocity of the ball can be calculated by,
Using v=u+at
∴ 0=u−10×3 ⟹u=30 m/s
Using S=ut+
2
1
at
2
∴ S=30×3+
2
1
×(−10)×3
2
=45m
Answer:
The travel would take 6.7 years.
Explanation:
The equation for an object moving in a straight line with acceleration is:
x = x0 + v0 t + 1/2a*t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
In a movement with constant speed, a = 0 and the equation for the position will be:
x = x0 + v t
where v = velocity
Let´s calculate the position from the Earth after half a year moving with an acceleration of 1.3 g = 1.3 * 9.8 m/s² = 12.74 m/s²:
Seconds in half a year:
1/2 year = 1.58 x 10⁷ s
x = 0 m + 0 m/s + 1/2 * 12.74 m/s² * (1.58 x 10⁷ s)² = 1.59 x 10¹⁵ m
Now let´s see how much time it takes the travel to the nearest star after this half year.
The velocity will be the final velocity achived after the half-year travel with an acceleration of 12.74 m/s²
v = v0 + a t
Since the spacecraft starts from rest, v0 = 0
v = 12.74 m/s² * 1.58 x 10⁷ s = 2.01 x 10 ⁸ m/s
Using the equation for position:
x = x0 + v t
4.1 x 10¹⁶ m = 1.59 x 10¹⁵ m + 2.01 x 10 ⁸ m/s * t
(4.1 x 10¹⁶ m - 1.59 x 10¹⁵ m) / 2.01 x 10 ⁸ m/s = t
t = 2.0 x 10⁸ s * 1 year / 3.2 x 10 ⁷ s = 6.2 years.
The travel to the nearest star would take 6.2 years + 0.5 years = 6.7 years.
